Problem: Is ${412260}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {412260}= &&{4}\cdot100000+ \\&&{1}\cdot10000+ \\&&{2}\cdot1000+ \\&&{2}\cdot100+ \\&&{6}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {412260}= &&{4}(99999+1)+ \\&&{1}(9999+1)+ \\&&{2}(999+1)+ \\&&{2}(99+1)+ \\&&{6}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {412260}= &&\gray{4\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {4}+{1}+{2}+{2}+{6}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${412260}$ is divisible by $3$ if ${ 4}+{1}+{2}+{2}+{6}+{0}$ is divisible by $3$ Add the digits of ${412260}$ $ {4}+{1}+{2}+{2}+{6}+{0} = {15} $ If ${15}$ is divisible by $3$ , then ${412260}$ must also be divisible by $3$ ${15}$ is divisible by $3$, therefore ${412260}$ must also be divisible by $3$.